To determine the time at which the particle's acceleration is 6 m/s², we need to follow these steps:
1. Displacement Function:
The displacement given is [tex]\( x(t) = 3t^2 + 2t + 1 \)[/tex], where [tex]\( x \)[/tex] is in meters and [tex]\( t \)[/tex] is in seconds.
2. First Derivative – Velocity:
The first derivative of [tex]\( x(t) \)[/tex] with respect to time [tex]\( t \)[/tex] gives us the velocity [tex]\( v(t) \)[/tex] of the particle.
[tex]\[
v(t) = \frac{dx}{dt} = \frac{d}{dt}(3t^2 + 2t + 1)
\][/tex]
Differentiating the displacement function:
[tex]\[
v(t) = 6t + 2
\][/tex]
3. Second Derivative – Acceleration:
The second derivative of [tex]\( x(t) \)[/tex] with respect to time [tex]\( t \)[/tex] gives us the acceleration [tex]\( a(t) \)[/tex] of the particle.
[tex]\[
a(t) = \frac{dv}{dt} = \frac{d}{dt}(6t + 2)
\][/tex]
Differentiating the velocity function:
[tex]\[
a(t) = 6
\][/tex]
4. Setting the Acceleration Equal to 6 m/s²:
The goal is to find the time [tex]\( t \)[/tex] when the acceleration [tex]\( a(t) \)[/tex] is 6 m/s².
[tex]\[
6 = 6
\][/tex]
Since the calculated acceleration [tex]\( a(t) \)[/tex] is a constant 6 m/s² for all times [tex]\( t \)[/tex], it means the acceleration does not depend on [tex]\( t \)[/tex]. Therefore, it is always 6 m/s² regardless of the value of [tex]\( t \)[/tex].
Thus, there is no specific time [tex]\( t \)[/tex] to solve for because the acceleration is constant and always equal to 6 m/s². Consequently, the set of solutions for [tex]\( t \)[/tex] when the acceleration is 6 m/s² is an empty set, indicating no specific time needs to be identified.
